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1 By shining light through samples, we can learn about the molecular structure of the sample and (sometimes) deduce its structure. IR UV RADIO By the end of this course you will be able to interpret these light absorbing properties of molecules and deduce molecular structures! HC H H H HC H H C H H H HC H H C H H C H H H HC H H C H H C H H C H H H HC H H C C H C H H H H H H C 1 H 4 C 2 H 6 C 3 H 8 C 4 H 10 C 4 H 10 Saturated alkanes (CH only compounds) have the general formula: C N H N+2 Note: it doesn’t matter how the atoms are configured:
22
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By shining light through samples,we can learn about the molecularstructure of the sample and(sometimes) deduce its structure.

IR

UV

RADIO

By the end of this course you willbe able to interpret these lightabsorbing properties of moleculesand deduce molecular structures!

H CH

HH

H CH

HCHH

H

H CH

HCH

HCH

HH

H CH

HCH

HCH

HCH

HH H C

H

HC

C

HCH

HH

HH

H

C1H4

C2H6

C3H8

C4H10C4H10

Saturated alkanes (CH onlycompounds) have the generalformula:

CNHN+2

Note: it doesn’t matter how theatoms are configured:

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H CH

HCH

HH

H CH

HCH

HCH

HH

C2H6

C3H8

CH

HCH

HCH

HHH C

H

HCH

HCH

HCH

H

C7H16

CNHN+2

C CC

H H

HH

HH

C CH

HH

H

C2H4

C3H6

CH

HCH

HCH

HHH C

H

HCH

HCH

CH

C7H14

CNHN

C CC

H H

HH

HC CH

C2H2

C3H6

C7H12

CCCCCC

H H HH

HH

HHCH

HH

HH

CNHN-2

Two less hydrogens per double bond or ring!

The molecular formula can tell you about the structure:FORMULA TELLS US THE NUMBER OF DOUBLE BONDS ORRINGS:

IHD = index of hydrogen deficiency.a.k.a. Degree of unsaturation.

For CNHM

IHD = 2N+2-M 2

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Example: C6H10

IHD = (6 x 2)+2-10 = 4 = 22 2

•Two double bonds

•Two rings

•One triple bonds

•One double bond plus one ring

C H

F

Cl

Br

I

O N

For each Carbon(valence =4) you add toa structure requires theaddition of two moreHydrogens

Each halogen(valance = 1),takes the place ofa hydrogen

Oxygen(valance = 2)can be addedinto a structurewithoutchanging thenumber ofhydrogensH C

H

HCHH

H

H CH

HCH

HCH

HH

C

Each addednitrogen(valence =3)requires oneadditionalhydrogenX =

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For CNHXXONIHD = 2N+2-X-(#halogens)+(#Ns) 2

C3H5BrO

Example: IHD =

C4H7NO

Example: IHD =

We can tell something about the molecules you have by thefrequencies of light it absorbs

CH3OH

C C H

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Different frequencies of light (electromagnetic radiation) interact withdifferent aspects molecular motion (potential and kinetic energy).

X-ray:core electronexcitation

UV:valanceelectronicexcitation

IR:molecularvibrations

Radio waves:Nuclear spin states(in a magnetic field)

General Principle of Spectroscopy: A molecule absorbs lightwhen the frequency of light correlates with an energy transition.

Molecules have quantized energy levels:ex. electronic energy levels.

ener

gy

ener

gy

hv

ener

gy } ΔΕ = hv

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Some (not all) nuclei have spin.

Nuclear spin can be examined by Nuclear Magnetic Resonance (NMR)

Only atoms with an odd number of protons or neutrons (or both) havenuclear spins that can be observed by NMR.

=

=En

ergy

In the absence of a magneticfield, both spin states haveequal energy

Ener

gy

No Field Magnetic Field StrongerMagnetic Field

hv

Ener

gy } ΔΕ = hvIn a strong magnetic field, the energylevel difference corresponds to theenergy of radio waves

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How an NMR spectrometerworks:

+ -

N S

RFtransmitter

RFReceiver

Note modernNMRs usesuperconductingmagnets to attainvery strongmagnetic fields

13C-NMRWe can examine the nuclear magnetic properties of carbon atomsin a molecule to learn about a molecules structure.

NOTE: most carbons are 12C. 12C has an even number of protonsand neutrons and cannot be observed by NMR techniques.However approximately 1% of carbons are 13C, and these we cansee in the NMR. (this will be come later when we considernuclear spin coupling)

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A 13C-NMR spectrum.

RF a

bsor

banc

e

RF Frequency

Here is what’s cool about 13C-NMR:

Different carbons appear at differentfrequencies! - Now you know howmany different types of Carbons youhave!

The intensity of the peak doesn’t does not necessarilycorrelate to the number of carbons.

An internalstandardcalled TMSis used toset thescale tozero

Example: Two alcohols (-OH compound) with formula C3H8O1. IHD = 02. Possible structures:

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F

OH

Cl

FF F

How Many 13C NMR resonances?

equivalent resonance structures

=1

BrBr

Br

Br

Br

Br

OH

Cl Br

Br

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An isolated 13C atoms have the same chemical shift. (resonate atthe same frequency)

CNucleus

Electron “cloud”

The electron cloud of the atom partially shields the nucleus from thesurrounding magnetic field

When the carbon nucleus is adjacent to an electronegative atom, thecarbon nuclei has fewer Electrons around it.

C F

Nuclei with less electron density around them will resonate at a differentfrequency: It requires shorter wavelength radio waves (or a weaker magneticfield) to cause a carbon nuclei to resonate if it has less electron density around it.

Conversely atoms that donate electron density cause nuclei to resonate atlower frequencies (higher field strengths).

Si

CH3

CH3

CH3CH3

The silicon in tetramethylsilane (TMS)shields the carbon nuclei and makes themappear “up-field”

The chemical shift spectrum ismeasured relative to TMS.

TMSShift

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TMSShift

δ= is used to signify chemical shift.Because the frequency depends on thefield strength,one uses a relative units scale of “partsper million” or ppm

Chemical shifts reported as ppm units give the same valuesfor the same compound regardless of the instrument used!

δ = (v(compound)- v(TMS))/ v(instrument field strength)

THE δ SCALE IS USED TO MEASURE CHEMICAL SHIFT

HCHHF

HCHHCl

HCHHBr

HCHHOH

HCHHCH3

7.3 ppm

50.2 ppm

71.6 ppm

25.6 ppm

9.6 ppm

HCHHCl

ClCHHCl

ClCClHCl

ClCClClCl

25.6 54.0 77.2 96.1

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H3CCH

CH2

H2C

CH2

CH3

OH

F

H2C

CH2

H2C

CH2

H2C

CH2

H2C

CH3

64

40 27

25

23

2284

30.6

25.3

29.3

29.3

31.9

22.7

14.1

Atoms which are not directly attached to an electronegativeatom are still shifted but the effect is significantly attenuated.The inductive effect (through bond influence) drops off rapidlywith distance.

C O C C C C C

050100150200

40 -5

sp3

90 70sp150 105

sp2

150 120

aromatic

210 160

carbonyl

Typical Values of Chemical shift depend on the1. Hybridization2. Electronegativity of attached atom(s).

C O

C Br, (Cl)

75 50

Some specific values are included in tables 12.1-12.9 in your text book!

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C4H6

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C3H6O

OH

13C peaks are in reality split by bonded protons.

Generallyshown like this:

Actually lookslike this:

HC

Why??

C H Adjacent nuclear spin of the proton locally increasesmagnetic field felt by carbon nuclei.

Carbons next toup proton spin

Carbons next todown proton spin

NOTE: There are onlytwo possible peaks forC’s having one adjacentproton. or

A “doublet”

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HC Appears as a doublet

HCH

Appears as a triplet

or

There are four possibilities which lead to two net spin states:

The distribution of spin states is random so half the time the carbon will feel no netspin, 25% two up spins and 25% of the time two down spins:

HCHH

Appears as quartet.

There are eight possibilities:

The general rule is:

The number of peaks observed is equal to the number of attached protons,(N), plus one.

Splitting = N+1

FOR 13C it is the number of protons directlyattached to the carbon that cause splitting.

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Why don’t the spins on adjacentCarbons split each other?

C COnly 1% of Carbons have a spin! (Most are 12C and are NMRinactive). Therefore only 1 in ten thousand pairs of Cs have bothhave spin and are coupled. That’s too small to detect!

Note all of these split peaks can quickly get confusing:

Am I looking at two different carbon resonances or a singledoublet?

For this reason we show our spectra without splitting. But we oftensee the splitting indicated over the peak:

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C6H10OIHD = ???

C3H9N44.6 (t)

27.4 (t)

11.5 (q)

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It is helpful to know what typical values of unfunctinalized C’s

Approximate Chemical Shifts ranges for unsubstituted Alkanes-CH3 5-22 ppm-CH2- 15-33 ppm>CH- 25-35

C=C 110-150

Alkyne 66-90

If your molecule contains Oxygen:

OCR H

OCR R

OCR O

OCR NHR

R = alkylOC

R X

X = Cl, Br, I

195 - 220doublet

195 - 220but singlet!

165-180 165-180 165-180

OC

H3C O

OC

H3C NH

60.4 34.4

OC

O O

~155

C OH

C O C

50 - 75 ppm 50 - 75ppm (but two of them!)

CO O

90 - 100 ppm

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If your molecule contains Nitrogen:

C NH2

28-47 ppm

CN

C

H

CN

C

CAmines:

C NO

O61 - 85 ppm

Nitro-

CN RImine

~155 ppm

C NC

14-30

~120

C3H7N139 (d)

113 (t)

44.8 (t)

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C3H9N43.0 (d)

26.5 (q)

C3H8O2

72.7 (d)

67.7 (t)

18.7 (q)

OHHO

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C6H8O

IHDSymmetryWhat is the O doing?What other groups can you identify?Can you account for the IHD?

C8H8OIHDSymmetryWhat is the O doing?What other groups can you identify?

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What You Should NowKnow!

• Bond Line Formulas• How to calculate IHDs• 13C chemical shift scale• Multiplicities

– s, d, t, q• Approximate Chemical Shifts

– Alkanes, Alkenes, Alkynes, Aromatics– Alcohol, Amines, Ethers, Carbonyls